In the figure, each inside square is formed by joining the midpoints of the sides of the next larger square. The area of the smallest shaded square is to be found. The outermost square has a side length of 10 cm. 
Step 1: Area of the outermost square.
Side = 10 cm → Area = \(10^2 = 100\) cm².
Step 2: Area ratio for midpoint-joined squares.
Each new square = \( \frac{1}{2} \) × area of previous square.
Thus areas form the sequence:
\[
100,\; 50,\; 25,\; 12.5,\; 6.25,\; 3.125,\; \dots
\]
Step 3: Identify the smallest shaded square.
According to the diagram, the smallest (innermost) shaded square corresponds to
\[
100 \times \left(\frac{1}{2}\right)^5 = 3.125.
\]
Final Answer: 3.125

In \(\triangle ABC\), \(DE \parallel BC\). If \(AE = (2x+1)\) cm, \(EC = 4\) cm, \(AD = (x+1)\) cm and \(DB = 3\) cm, then the value of \(x\) is

In the adjoining figure, PA and PB are tangents to a circle with centre O such that $\angle P = 90^\circ$. If $AB = 3\sqrt{2}$ cm, then the diameter of the circle is
In the adjoining figure, TS is a tangent to a circle with centre O. The value of $2x^\circ$ is
A continuous time periodic signal \( x(t) \) is given by: \[ x(t) = 1 + 2\cos(2\pi t) + 2\cos(4\pi t) + 2\cos(6\pi t) \] If \( T \) is the period of \( x(t) \), then evaluate: \[ \frac{1}{T} \int_0^T |x(t)|^2 \, dt \quad {(round off to the nearest integer).} \]
The maximum percentage error in the equivalent resistance of two parallel connected resistors of 100 \( \Omega \) and 900 \( \Omega \), with each having a maximum 5% error, is: \[ {(round off to nearest integer value).} \]
Consider a distribution feeder, with \( R/X \) ratio of 5. At the receiving end, a 350 kVA load is connected. The maximum voltage drop will occur from the sending end to the receiving end, when the power factor of the load is: \[ {(round off to three decimal places).} \]
In the circuit with ideal devices, the power MOSFET is operated with a duty cycle of 0.4 in a switching cycle with \( I = 10 \, {A} \) and \( V = 15 \, {V} \). The power delivered by the current source, in W, is: \[ {(round off to the nearest integer).} \] 
The induced emf in a 3.3 kV, 4-pole, 3-phase star-connected synchronous motor is considered to be equal and in phase with the terminal voltage under no-load condition. On application of a mechanical load, the induced emf phasor is deflected by an angle of \( 2^\circ \) mechanical with respect to the terminal voltage phasor. If the synchronous reactance is \( 2 \, \Omega \), and stator resistance is negligible, then the motor armature current magnitude, in amperes, during loaded condition is closest to: \[ {(round off to two decimal places).} \]